CAIE FP1 2019 November — Question 11 EITHER 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeSolve via substitution then back-substitute
DifficultyChallenging +1.8 This Further Maths question requires multiple sophisticated steps: deriving derivatives of w from y using chain rule, algebraic manipulation with trigonometric identities to verify the transformed DE, solving a second-order linear DE with particular integral, then back-substituting through inverse trig. The verification in part (i) demands careful algebraic skill, and converting initial conditions between variables adds complexity. Significantly harder than standard A-level but represents expected FM difficulty rather than exceptional challenge.
Spec4.10c Integrating factor: first order equations4.10e Second order non-homogeneous: complementary + particular integral

It is given that \(w = \cos y\) and $$\tan y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } + 2 \tan y \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1 + \mathrm { e } ^ { - 2 x } \sec y$$
  1. Show that $$\frac { \mathrm { d } ^ { 2 } w } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} w } { \mathrm {~d} x } + w = - \mathrm { e } ^ { - 2 x }$$
  2. Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , y = \frac { 1 } { 3 } \pi\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 3 } }\). [10]

It is given that $w = \cos y$ and

$$\tan y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } + 2 \tan y \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1 + \mathrm { e } ^ { - 2 x } \sec y$$

(i) Show that

$$\frac { \mathrm { d } ^ { 2 } w } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} w } { \mathrm {~d} x } + w = - \mathrm { e } ^ { - 2 x }$$

(ii) Find the particular solution for $y$ in terms of $x$, given that when $x = 0 , y = \frac { 1 } { 3 } \pi$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 3 } }$. [10]\\

\hfill \mbox{\textit{CAIE FP1 2019 Q11 EITHER [10]}}